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Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?
\((20 \times 16)\) women can complete the work in 1 day.
\(\therefore \quad 1\) woman’s 1 day’s work \(=\frac{1}{320}\).
\((16 \times 15)\) men can complete the work in 1 day.
\(\therefore \quad 1\) man’s 1 day’s work \(=\frac{1}{240}\).
So, required ratio \(=\frac{1}{240}: \frac{1}{320}=4: 3\).
10 men can complete a piece of work in 15 days and 15 women can complete the same work in 12 days. If all 10 men and 15 women work together, in how many days will the work get completed?
\(
\begin{aligned}
& 10 \text { men’s } 1 \text { day’s work }=\frac{1}{15} \\
& 15 \text { women’s } 1 \text { day’s work }=\frac{1}{12} . \\
& \left(10 \text { men }+15 \text { women)’s } 1 \text { day’s work }=\left(\frac{1}{15}+\frac{1}{12}\right)=\frac{9}{60}=\frac{3}{20} .\right.
\end{aligned}
\)
\(\therefore \quad 10\) men and 15 women will complete the work in \(\frac{20}{3}=6 \frac{2}{3}\) days.
A job can be done by 3 skilled worksmen in 20 days or by 5 boys in 30 days. How many days will they take if they work together? (M.A.T., 2009)
\(
\begin{aligned}
& 3 \text { men’s } 1 \text { day’s work }=\frac{1}{20} \\
& 5 \text { boys’ } 1 \text { day’s work }=\frac{1}{30} . \\
& (3 \text { men }+5 \text { boys }) \text { ‘s } 1 \text { day’s work }=\left(\frac{1}{20}+\frac{1}{30}\right)=\frac{5}{60}=\frac{1}{12} .
\end{aligned}
\)
\(\therefore \quad 3\) men and 5 boys will complete the work in 12 days.
Five men are working to complete a work in 15 days. After five days 10 women are accompanied by them to complete the work in next 5 days. If the work is to be done by women only, then in how many days could the work be over if 10 women have started it ? (Bank Recruitment, 2007)
\(
\begin{aligned}
& 5 \text { men’s } 15 \text { days’ work }=5 \text { men’s } 10 \text { days’ work }+10 \\
& \text { women’s } 5 \text { days’ work } \\
& \Rightarrow \quad 5 \text { men’s } 5 \text { days’ work }=10 \text { women’s } 5 \text { days’ work } \\
& \Rightarrow \quad 10 \text { women’s } 5 \text { days’ work }=\left(\frac{1}{15} \times 5\right)=\frac{1}{3} \\
& \Rightarrow \quad 10 \text { women’s } 1 \text { day’s work }=\frac{1}{15} \\
& \therefore \quad 10 \text { women can complete the work in } 15 \text { days. }
\end{aligned}
\)
A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in the stipulated time. If he had not engaged the additional men, how many days behind the schedule the work should have been finished? (G.B.O., 2007)
100 men’s 40 days’ work +100 men’s 5 days’ work \(=1\) \(\Rightarrow \quad 100\) men’s 45 days’ work \(=1\)
So, if the contractor had not engaged additional men, 100 men would have finished the work in 45 days.
Difference in time \(=(45-40)\) days \(=5\) days.
Seven men can complete a work in 12 days. They started the work and after 5 days, two men left. In how many days will the work be completed by the remaining men?
\((7 \times 12)\) men can complete the work in 1 day.
\(
\begin{aligned}
& \therefore 1 \text { man’s } 1 \text { day’s work }=\frac{1}{84} . \\
& 7 \text { men’s } 5 \text { days’ work }=\left(\frac{1}{12} \times 5\right)=\frac{5}{12} . \\
& \text { Remaining work }=\left(1-\frac{5}{12}\right)=\frac{7}{12} . \\
& 5 \text { men’s } 1 \text { day’s work }=\left(\frac{1}{84} \times 5\right)=\frac{5}{84} .
\end{aligned}
\)
\(\frac{5}{84}\) work is done by them in 1 day.
\(\frac{7}{12}\) work is done by themin \(\left(\frac{84}{5} \times \frac{7}{12}\right)=\frac{49}{5}\) days \(=9 \frac{4}{5}\) days.
12 men complete a work in 9 days. After they have worked for 6 days, 6 more men join them. How many days will they take to complete the remaining work?
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{108} \\
& 12 \text { men’s } 6 \text { days’ work }=\left(\frac{1}{9} \times 6\right)=\frac{2}{3} \\
& \text { Remaining work }=\left(1-\frac{2}{3}\right)=\frac{1}{3} \\
& 18 \text { men’s } 1 \text { day’s work }=\left(\frac{1}{108} \times 18\right)=\frac{1}{6}
\end{aligned}
\)
\(\frac{1}{6}\) work is done by them in 1 day.
\(\therefore \quad \frac{1}{3}\) work is done by them in \(\left(6 \times \frac{1}{3}\right)=2\) days.
Three men, four women and six children can complete a work in seven days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?
Let 1 woman’s 1 day’s work \(=x\).
Then, 1 man’s 1 day’s work \(=\frac{x}{2}\) and 1 child’s 1 day’s work \(=\frac{x}{4}\)
So, \(\left(\frac{3 x}{2}+4 x+\frac{6 x}{4}\right)=\frac{1}{7} \Rightarrow \frac{28 x}{4}=\frac{1}{7}\)
\(
\Rightarrow x=\left(\frac{1}{7} \times \frac{4}{28}\right)=\frac{1}{49} \text {. }
\)
\(\therefore 1\) woman alone can complete the work in 49 days. So, to complete the work in 7 days, the number of women required \(=\left(\frac{49}{7}\right)=7\)
A man, a woman and a boy can complete a job in 3,4 and 12 days respectively. How many boys must assist 1 man and 1 woman to complete the job in \(\frac{1}{4}\) of a day?
\((1\) man +1 woman \()\) ‘s 1 day’s work \(=\left(\frac{1}{3}+\frac{1}{4}\right)=\frac{7}{12}\)
Work done by 1 man and 1 woman in \(\frac{1}{4}\) day \(=\left(\frac{7}{12} \times \frac{1}{4}\right)=\frac{7}{48}\).
Remaining work \(=\left(1-\frac{7}{48}\right)=\frac{41}{48}\).
Work done by 1 boy in \(\frac{1}{4}\) day \(=\left(\frac{1}{12} \times \frac{1}{4}\right)=\frac{1}{48}\).
\(\therefore \quad\) Number of boys required \(=\left(\frac{41}{48} \times 48\right)=41\).
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same
work?
1 man’s 1 day’s work \(=\frac{1}{100}\).
\((10\) men +15 women \()\) ‘s 1 day’s work \(=\frac{1}{6}\).
15 women’s 1 day’s work \(=\left(\frac{1}{6}-\frac{10}{100}\right)=\left(\frac{1}{6}-\frac{1}{10}\right)=\frac{1}{15}\).
1 woman’s 1 day’s work \(=\frac{1}{225}\).
\(\therefore \quad 1\) woman alone can complete the work in 225 days.
A child can do a piece of work 15 hours slower than a woman. The child works for 18 hours on the job and then the woman takes charge for 6 hours. In this manner, \(\frac{3}{5}\) of the work can be completed. To complete the job now, how much time will the woman take? (M.A.T., 2005)
Suppose the woman takes \(x\) hours to do the job. Then, the child takes \((x+15)\) hours to do the job.
Woman’s 1 hours’ work \(=\frac{1}{x}\).
Child’s 1 hours’ work \(=\frac{1}{(x+15)}\).
Child’s 18 hours’ work + Woman’s 6 hours’ work \(=\frac{3}{5}\)
\(\Rightarrow \quad \frac{18}{(x+15)}+\frac{6}{x}=\frac{3}{5} \Rightarrow \frac{18 x+6(x+15)}{x(x+15)}=\frac{3}{5}\)
\(\Rightarrow \quad 5(24 x+90)=3\left(x^2+15 x\right)\)
\(\Rightarrow \quad 120 x+450=3 x^2+45 x\)
\(\Rightarrow \quad 3 x^2-75 x-450=0 \Rightarrow x^2-25 x-150=0\)
\(\Rightarrow \quad x^2-30 x+5 x-150=0\)
\(\Rightarrow \quad x(x-30)+5(x-30)=0\)
\(\Rightarrow \quad(x-30)(x+5)=0 \Rightarrow x=30\).
Remaining work \(=\left(1-\frac{3}{5}\right)=\frac{2}{5}\).
\(\frac{1}{30}\) work is done by the woman in 1 hour.
\(\therefore \quad \frac{2}{5}\) work will be done by the woman in \(\left(30 \times \frac{2}{5}\right)=12 \text { hours. }\)
A group of workers having equal efficiency can complete a job in 4 days. But it so happened that every alternate day starting from the second day, 3 workers are withdrawn from the job and every alternate day starting from the third day, 2 workers are added to the group. If it now takes 7 days to complete the job, find the number of workers who started the job.
Let the number of workers who started the job be \(n\). Then, \(n\) workers’ 1 day’s work \(=\frac{1}{4}\) \(\Rightarrow \quad 1\) worker’s 1 day’s work \(=\frac{1}{4 n}\).
Now, \(n\) workers worked on first day, \((n-3)\) on 2 nd day, \((n-3+2)\) i.e., \((n-1)\) on \(3 r d\) day, and so on. Thus, we have:
\(
\begin{aligned}
& {[n+(n-3)+(n-1)+(n-4)+(n-2)+(n-5)+(n} \\
& -3)] \times \frac{1}{4 n}=1 \\
& \Rightarrow 7 n-18=4 n \Rightarrow 3 n=18 \Rightarrow n=6 .
\end{aligned}
\)
Hence, 6 workers started the job.
A man, a woman and a boy can do a piece of work in 6, 9 and 18 days respectively. How many boys must assist one man and one woman to do the work in 1 day? (N.M.A.T., 2006)
\(
\begin{aligned}
& (1 \text { man }+1 \text { woman })^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{6}+\frac{1}{9}=\frac{5}{18} \\
& \text { Remaining work }=\left(1-\frac{5}{18}\right)=\frac{13}{18} .
\end{aligned}
\)
Work done by 1 boy in 1 day \(=\frac{1}{18}\).
\(\therefore \quad\) Number of boys required \(=\left(\frac{13}{18} \times 18\right)=13\).
If 3 men or 9 boys can finish a piece of work in 21 days, in how many days can 5 men and 6 boys together do the same piece of work? (Bank Recruitment, 2010)
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{21 \times 3}=\frac{1}{63} \\
& 1 \text { boy’s } 1 \text { day’s work }=\frac{1}{21 \times 9}=\frac{1}{189} \text {. } \\
& (5 \text { men }+6 \text { boys)’s } 1 \text { day’s work } \\
& =\frac{5}{63}+\frac{6}{189}=\frac{5}{63}+\frac{2}{63}=\frac{7}{63}=\frac{1}{9} \text {. } \\
&
\end{aligned}
\)
Hence, 5 men and 6 boys together can do the work in 9 days.
If 2 men or 6 women or 4 boys can finish a work in 99 days, how many days will one man, one woman and one boy together take to finish the same work? (Bank Recruitment, 2010)
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{99 \times 2}=\frac{1}{198} \\
& 1 \text { woman’s } 1 \text { day’s work }=\frac{1}{99 \times 6}=\frac{1}{594} \\
& 1 \text { boy’s } 1 \text { day’s work }=\frac{1}{99 \times 4}=\frac{1}{396} \\
& \begin{array}{l}
(1 \text { man }+1 \text { woman }+1 \text { boy)’s } 1 \text { day’s work } \\
\quad=\left(\frac{1}{198}+\frac{1}{594}+\frac{1}{396}\right)=\frac{11}{1188}=\frac{1}{108}
\end{array}
\end{aligned}
\)
Hence, 1 man, 1 woman and 1 boy together take 108 days to finish the same work.
8 men can complete a piece of work in 20 days. 8 women can complete the same work in 32 days. In how many days will 5 men and 8 women together complete the same work? (Bank P.O., 2010)
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{20 \times 8}=\frac{1}{160} \\
& 1 \text { woman’s } 1 \text { day’s work }=\frac{1}{32 \times 8}=\frac{1}{256} . \\
& (5 \text { men }+8 \text { women }) \text { ‘s } 1 \text { day’s work } \\
& \quad=\left(\frac{5}{160}+\frac{8}{256}\right)=\frac{1}{32}+\frac{1}{32}=\frac{1}{16} .
\end{aligned}
\)
Hence, 5 men and 8 women together can complete the work in 16 days.
18 men can complete a piece of work in 63 days. 9 women take 189 days to complete the same piece of work. How many days will 4 men, 9 women and 12 children together take to complete the piece of work if 7 children alone can complete the piece of work in 486 days? (I.R.M.A., 2007)
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{63 \times 18}=\frac{1}{1134} \\
& 1 \text { woman’s } 1 \text { day’s work }=\frac{1}{189 \times 9}=\frac{1}{1701} \\
& 1 \text { child’s } 1 \text { day’s work }=\frac{1}{486 \times 7}=\frac{1}{3402} \\
& (4 \text { men }+9 \text { women }+12 \text { children }) \text { ‘s } 1 \text { days’ work } \\
& =\left(\frac{4}{1134}+\frac{9}{1701}+\frac{12}{3402}\right)=\frac{42}{3402}=\frac{1}{81}
\end{aligned}
\)
Hence, 4 men, 9 women and 12 children together will complete the work in 81 days.
16 men can finish a work in 24 days and 48 boys can finish the same work in 16 days. 12 men started the work and after 4 days 12 boys joined them. In how many days can they finish the remaining work? (R.R.B., 2008)
1 man’s 1 day’s work \(=\frac{1}{24 \times 16}=\frac{1}{384}\)
1 boy’s 1 day’s work \(=\frac{1}{16 \times 48}=\frac{1}{768}\).
12 men’s 4 days’ work \(=\left(\frac{12}{384} \times 4\right)=\frac{1}{8}\).
Remaining work \(=\left(1-\frac{1}{8}\right)=\frac{7}{8}\).
(12 men +12 boys)’s 1 day’s work
\(
=\left(\frac{12}{384}+\frac{12}{768}\right)=\left(\frac{1}{32}+\frac{1}{64}\right)=\frac{3}{64} \text {. }
\)
\(\frac{3}{64}\) work is done by ( 12 men +12 boys \()\) in 1 day.
\(\therefore \quad \frac{7}{8}\) work is done by them in \(\left(\frac{64}{3} \times \frac{7}{8}\right)=\frac{56}{3}\) days \(=18 \frac{2}{3}\) days.
12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days, all of them stopped working. How many women should be put on the job to complete the remaining work, if it is to be completed in 3 days?
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{48} \\
& 1 \text { woman’s } 1 \text { day’s work }=\frac{1}{60} . \\
& 6 \text { men’s } 2 \text { days’ work }=\left(\frac{6}{48} \times 2\right)=\frac{1}{4} . \\
& \text { Remaining work }=\left(1-\frac{1}{4}\right)=\frac{3}{4} .
\end{aligned}
\)
Now, \(\frac{1}{60}\) work is done in 1 day by 1 woman.
So, \(\frac{3}{4}\) work will be done in 3 days by \(\left(60 \times \frac{3}{4} \times \frac{1}{3}\right)\) \(=15\) women
Twelve children take sixteen days to complete a work which can be completed by eight adults in twelve days. Sixteen adults started working and after three days ten adults left and four children joined them. How many days will they take to complete the remaining work?
\(
\begin{aligned}
& 1 \text { child’s } 1 \text { day’s work }=\frac{1}{192} ; \\
& 1 \text { adult’s } 1 \text { day’s work }=\frac{1}{96} . \\
& \text { Work done in } 3 \text { days }=\left(\frac{1}{96} \times 16 \times 3\right)=\frac{1}{2} . \\
& \text { Remaining work }=\left(1-\frac{1}{2}\right)=\frac{1}{2} .
\end{aligned}
\)
\(\left(6\right.\) adults +4 children)’s 1 day’s work \(=\left(\frac{6}{96}+\frac{4}{192}\right)=\frac{1}{12}\). \(\frac{1}{12}\) work is done by them in 1 day.
\(\frac{1}{2}\) work is done by them in \(\left(12 \times \frac{1}{2}\right)=6\) days.
Sixteen men can complete a work in twelve days. Twenty-four children can complete the same work in eighteen days. Twelve men and eight children started working and after eight days three more children joined them. How many days will they now take to complete the remaining work?
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{192} \\
& 1 \text { child’s } 1 \text { day’s work }=\frac{1}{432} \\
& \text { Work done in } 8 \text { days }=8\left(\frac{12}{192}+\frac{8}{432}\right)=8\left(\frac{1}{16}+\frac{1}{54}\right)=\frac{35}{54} \\
& \text { Remaining work }=\left(1-\frac{35}{54}\right)=\frac{19}{54} \\
& (12 \text { men }+11 \text { children }) \text { ‘s } 1 \text { day’s work }=\left(\frac{12}{192}+\frac{11}{432}\right)=\frac{19}{216}
\end{aligned}
\)
Now, \(\frac{19}{216}\) work is done by them in 1 day.
\(\therefore \quad \frac{19}{54}\) work will be donebythemin \(\left(\frac{216}{19} \times \frac{19}{54}\right)=4\) days.
Twenty-four men can complete a work in sixteen days. Thirty-two women can complete the same work in twenty-four days. Sixteen men and sixteen women started working and worked for twelve days. How many more men are to be added to complete the remaining work in 2 days?
1 man’s 1 day’s work \(=\frac{1}{384}\)
1 woman’s 1 day’s work \(=\frac{1}{768}\).
Work done in 12 days \(=12\left(\frac{16}{384}+\frac{16}{768}\right)=\left(12 \times \frac{3}{48}\right)=\frac{3}{4}\).
Remaining work \(=\left(1-\frac{3}{4}\right)=\frac{1}{4}\).
(16 men +16 women)’s 2 days’ work
\(
=2\left(\frac{16}{384}+\frac{16}{768}\right)=\left(2 \times \frac{1}{16}\right)=\frac{1}{8} \text {. }
\)
Remaining work \(=\left(\frac{1}{4}-\frac{1}{8}\right)=\frac{1}{8}\).
\(\frac{1}{384}\) work is done in 1 day by 1 man.
\(\therefore \quad \frac{1}{8}\) work will be done in 2 days by \(\left(384 \times \frac{1}{8} \times \frac{1}{2}\right)\)
\(=24 \mathrm{men}\)
5 men and 2 boys working together can do four times as much work as a man and a boy. Working capacities of a man and a boy are in the ratio:
Let 1 man’s 1 day’s work \(=x\) and 1 boy’s 1 day’s work \(=y\).
Then, \(\quad 5 x+2 y=4(x+y) \Rightarrow x=2 y \Rightarrow \frac{x}{y}=\frac{2}{1}\).
If 12 men and 16 boys can do a piece of work in 5 days; 13 men and 24 boys can do it in 4 days, then the ratio of the daily work done by a man to that of a boy is
Let 1 man’s 1 day’s work \(=x\) and 1 boy’s 1 day’s work \(=y\). Then, \(12 x+16 y=\frac{1}{5}\) and \(13 x+24 y=\frac{1}{4}\). Solving these two equations, we get: \(x=\frac{1}{100}\) and \(y=\frac{1}{200}\).
\(
\therefore \quad \text { Required ratio }=x: y=\frac{1}{100}: \frac{1}{200}=2: 1 .
\)
4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?
Let 1 man’s 1 day’s work \(=x\)
and 1 woman’s 1 day’s work \(=y\).
Then, \(4 x+6 y=\frac{1}{8}\) and \(3 x+7 y=\frac{1}{10}\).
Solving these two equations, we get : \(x=\frac{11}{400}, y=\frac{1}{400}\).
\(\therefore \quad 1\) woman’s 1 day’s work \(=\frac{1}{400}\).
\(\Rightarrow \quad 10\) women’s 1 day’s work \(=\left(\frac{1}{400} \times 10\right)=\frac{1}{40}\).
Hence, 10 women will complete the work in 40 days.
4 men and 10 women were put on a work. They completed \(\frac{1}{3}\) of the work in 4 days. After this 2 men and 2 women were increased. They completed \(\frac{2}{9}\) more of the work in 2 days. If the remaining work is to be completed in 3 days, then how many more women must be increased? (M.A.T., 2006)
Let 1 man’s 1 day’s work \(=x\)
and 1 woman’s 1 day’s work \(=y\).
Then, \(4 x+10 y=\frac{1}{3} \times \frac{1}{4}=\frac{1}{12} \Rightarrow 2 x+5 y=\frac{1}{24} \dots(i)\)
And, \(6 x+12 y=\frac{1}{9} \Rightarrow 2 x+4 y=\frac{1}{27} \dots(ii)\)
Subtracting (ii) from (i), we get: \(y=\frac{1}{24}-\frac{1}{27}=\frac{1}{216}\).
Now,\((6 \text { men }+12 \text { women })^{\prime}\) s 3 days \({ }^{\prime}\) work \(=\left(\frac{1}{9} \times 3\right)=\frac{1}{3}\).
Work completed \(=\left(\frac{1}{3}+\frac{2}{9}+\frac{1}{3}\right)=\frac{8}{9}\).
Remaining work \(=\left(1-\frac{8}{9}\right)=\frac{1}{9}\).
1 woman’s 3 days’ work \(=\left(\frac{1}{216} \times 3\right)=\frac{1}{72}\).
In 3 days, \(\frac{1}{72}\) work is done by 1 woman.
\(\therefore \quad\) In 3 days, \(\frac{1}{9}\) work is done by \(\left(72 \times \frac{1}{9}\right)=8\) women.
One man, 3 women and 4 boys can do a piece of work in 96 hours, 2 men and 8 boys can do it in 80 hours, 2 men and 3 women can do it in 120 hours. 5 men and 12 boys can do it in:
Let 1 man’s 1 hour’s work \(=x\);
1 woman’s 1 hour’s work \(=y\)
and 1 boy’s 1 hour’s work \(=z\).
Then, \(x+3 y+4 z=\frac{1}{96} \dots(i)\)
\(
\begin{aligned}
& 2 x+8 z=\frac{1}{80} \dots(ii)\\
& 2 x+3 y=\frac{1}{120} \dots(iii)
\end{aligned}
\)
Adding (ii) and (iii) and subtracting (i) from it,
we get: \(3 x+4 z=\frac{1}{96} \dots(iv)\)
From (ii) and (iv), we get \(x=\frac{1}{480}\).
Substituting, we get : \(y=\frac{1}{720}, z=\frac{1}{960}\).
\(
\begin{aligned}
& (5 \text { men }+12 \text { boys)’s } 1 \text { hour’s work } \\
& =\left(\frac{5}{480}+\frac{12}{960}\right)=\left(\frac{1}{96}+\frac{1}{80}\right)=\frac{11}{480} .
\end{aligned}
\)
\(\therefore \quad 5\) men and 12 boys can do the work in \(\frac{480}{11}\) i.e., \(43 \frac{7}{11}\) hours.
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be
Let 1 man’s 1 day’s work \(=x\)
and 1 boy’s 1 day’s work \(=y\).
Then, \(6 x+8 y=\frac{1}{10}\) and \(26 x+48 y=\frac{1}{2}\).
Solving these two equations, we get: \(x=\frac{1}{100}\) and \(y=\frac{1}{200}\).
\(\left(15\right.\) men +20 boys)’s 1 day’s work \(=\left(\frac{15}{100}+\frac{20}{200}\right)=\frac{1}{4}\).
\(\therefore \quad 15\) men and 20 boys can do the work in 4 days.
If 5 men and 3 women can reap 18 acre of crop in 4 days; 3 men and 2 women can reap 22 acre of crop in 8 days, then how many men are required to join 21 women to reap 54 acre of crop in 6 days? (M.C.A., 2005)
Acreage reaped by 5 men and 3 women in 1 day \(=\frac{18}{4}=\frac{9}{2}\). Acreage reaped by 3 men and 2 women in 1 day \(=\frac{22}{8}=\frac{11}{4}\).
Suppose 1 man can reap \(x\) acres in 1 day and 1 woman can reap \(y\) acres in 1 day.
\(
\begin{aligned}
& \therefore \quad 5 x+3 y=\frac{9}{2} \Rightarrow 10 x+6 y=9 \dots(i) \\
& 3 x+2 y=\frac{11}{4} \Rightarrow 9 x+6 y=\frac{33}{4} \dots(ii)
\end{aligned}
\)
Subtracting (ii) from (i), we get : \(x=9-\frac{33}{4}=\frac{3}{4}\).
Putting \(x=\frac{3}{4}\) in (i), we get: \(6 y=9-\frac{15}{2}=\frac{3}{2} \Rightarrow y=\frac{1}{4}\).
Acreage reaped by 21 women in 6 days \(=\left(\frac{1}{4} \times 21 \times 6\right)=\frac{63}{2}\).
Remaining acreage to be reaped \(=\left(54-\frac{63}{2}\right)=\frac{45}{2}\).
Acreage reaped by 1 man in 6 days \(=\left(\frac{3}{4} \times 6\right)=\frac{9}{2}\).
In 6 days, \(\frac{9}{2}\) acre is reaped by 1 man.
\(\therefore\) In 6 days, \(\frac{45}{2}\) acre is reaped by \(\left(\frac{2}{9} \times \frac{45}{2}\right)\) men \(=5\) men.
25 men with 10 boys can do in 6 days as much work as 21 men with 30 boys can do in 5 days. How many boys must help 40 men to do the same work in 4 days? (R.R.B., 2008)
Let 1 man’s 1 day’s work \(=x\)
and 1 boy’s 1 day’s work \(=y\).
Then, \(6(25 x+10 y)=5(21 x+30 y)\)
\(
\Rightarrow \quad 150 x+60 y=105 x+150 y \Rightarrow 45 x=90 y \Rightarrow x=2 y \text {. }
\)
Let the required number of boys be \(z\).
Then, \(4(40 x+z y)=6(25 x+10 y) \Rightarrow 4(80 y+z y)\)
\(
\begin{aligned}
& =6(50 y+10 y) \quad[\because x=2 y] \\
& \Rightarrow \quad 80+z=\frac{6 \times 60}{4}=90 \Rightarrow z=10 \\
&
\end{aligned}
\)
40 men can complete a piece of work in 15 days. 20 more men join them after 5 days they start doing work. How many days will be required by them to finish the remaining work? [ESIC—UDC, Exam 2016]
Work done by 40 men in 5 days \(=\frac{1}{3}\) (as if whole work is completed in 15 days then in 5 days \(1 / 3^{\text {rd }}\) of the work will be finished)
Remaining work \(=1-\frac{1}{3}=\frac{2}{3}\)
\(\because 40\) men do 1 work in 15 days.
60 men can do \(\frac{2}{3}\) work in \(x\) day
\(
\begin{aligned}
& \frac{M_1 D_1}{W_1}=\frac{M_2 D_2}{W_2} \\
& M_1=40 \quad M_2=60 \\
& D_1=15 \quad D_2=x \\
& W_1=1 \quad W_2=\frac{2}{3} \\
& \Rightarrow \frac{40 \times 15}{1}=\frac{60 \times x}{\frac{2}{3}} \\
& \Rightarrow \frac{2}{3}(40 \times 15)=60 x \\
& \Rightarrow 2 \times 40 \times 5=60 x \\
& \Rightarrow x=\frac{20}{3}=6 \frac{2}{3} \text { days }
\end{aligned}
\)
12 men can do a piece of work in 24 days. How many days are needed to complete the work, if 8 men do this work? [Indian Railway Gr. ‘D’ Exam, 2014]
12 men can do a piece of work in 24 days \(\Rightarrow M_1=12\) and \(D_1=24\)
8 men can do this work in \(D_2\) days
\(
\begin{aligned}
\Rightarrow & M_2=8 \\
& M_1 D_1=M_2 D_2 \\
\Rightarrow & 12 \times 24=8 \times D_2 \\
\Rightarrow & D_2=\frac{12 \times 24}{8}=36 \text { days }
\end{aligned}
\)
A can do in one day three times the work done by B in one day. They together finish \(\frac{2}{5}\) of the work in 9 days. The number of days by which \(B\) can do the work alone is [SSC-CHSL (10+2) Exam, 2015]
Let time taken by A alone in doing work be \(x\) days.
\(\therefore\) Time taken by B alone \(=3 x\) days
A’s 1 day’s work \(=\frac{1}{x}\)
B’s 1 day’s work \(=\frac{1}{3 x}\)
\(\because A\) and \(B\) together finish \(=\frac{2}{5}\) work in 9 days.
\(\therefore\) Time taken by A and B in doing whole work \(=\frac{9 \times 5}{2}=\frac{45}{2}\) days
According to given information we get
\(
\begin{aligned}
& \therefore \frac{1}{x}+\frac{1}{3 x}=\frac{2}{45} \\
& \Rightarrow \frac{3+1}{3 x}=\frac{2}{45} \\
& \Rightarrow \frac{4}{3 x}=\frac{2}{45}
\end{aligned}
\)
By cross-multiply we get
\(
\begin{aligned}
& \Rightarrow 2 \times 3 x=4 \times 45 \\
& \Rightarrow x=\frac{4 \times 45}{2 \times 3}=30 \text { days }
\end{aligned}
\)
Time taken by \(\mathrm{A}=x\) days \(=30\) days
\(\therefore\) Time taken by \(B=3 x\) days \(=3 \times 30=90\) days
A, B and C can complete a piece of work in 24, 5 and 12 days respectively. Working together, they will complete the same work in [SSC—CHSL (10+2) Exam, 2015]
A’s 1 days work \(=\frac{1}{24}\)
B’s 1 days work \(=\frac{1}{5}\)
C’s 1 days work \(=\frac{1}{12}\)
\(\therefore(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime}\) s 1 days work
\(=\frac{1}{24}+\frac{1}{5}+\frac{1}{12}\)
\(\mathrm{LCM}\) of 24,5 and 12
\(
\begin{array}{l|l}
2 & 24-5-12 \\
\hline 2 & 12-5-6 \\
\hline 3 & 6-5-3 \\
\hline & 2-5-1
\end{array}
\)
\(
\begin{aligned}
& 2 \times 2 \times 3 \times 2 \times 5=120 \\
& =\frac{5+24+10}{120} \\
& =\frac{39}{120}=\frac{13}{40}
\end{aligned}
\)
Time taken by A, B and C to complete the work, working together
\(
=\frac{40}{13}=3 \frac{1}{13} \text { days }
\)
\(X\) can do a piece of work in 24 days. When he had worked for 4 days, \(Y\) joined him. If complete work was finished in 16 days, \(Y\) can alone finish that work in [SSC-CHSL (10+2) Exam, 2015]
\(
\begin{aligned}
& X^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{24} \\
& X^{\prime} \text { s } 16 \text { day’s work }=\frac{16}{24}
\end{aligned}
\)
Let \(Y\) alone complete the work in \(x\) days.
\(
\text { Y’s } 12 \text { days work }=\frac{12}{x}
\)
According to the question,
Complete work done by \(X\) and \(Y=1\)
\(X^{\prime}\) s 16 days work \(+Y^{\prime}\) s 12 days work \(=1\)
\(
\begin{aligned}
& \Rightarrow \frac{16}{24}+\frac{12}{x}=1 \\
& \Rightarrow \frac{2}{3}+\frac{12}{x}=1 \\
& \Rightarrow \frac{12}{x}=1-\frac{2}{3}=\frac{1}{3} \\
& \Rightarrow x=12 \times 3=36 \text { days }
\end{aligned}
\)
6 men can complete a piece of work in 12 days, 8 women can complete the same piece of work in 18 days and 18 children can do it in 10 days. 4 men, 12 women and 20 children do the work for 2 days. If the remaining work be completed by men only in 1 day, how many men will be required? [RBI Officer Gr. ‘B’ (Phase-1) Online Exam, 2015]
6 men will complete the work in 12 days
1 man will complete the work in \(=6 \times 12=72\) days
8 women can complete two work in 18 day
1 woman will complete the work in \(=8 \times 18=144\) days 18 children can complete the work in 10 days.
1 child will complete the work in \(=18 \times 10=180\) days
1 men’s 1 day’s work \(=\frac{1}{72}\)
1 women’s 1 day’s work \(=\frac{1}{144}\)
1 children’s 1 day’s work \(=\frac{1}{180}\)
4 men +12 women +20 children’s 2 days’ work
\(=2\left(\frac{4}{72}+\frac{12}{144}+\frac{20}{180}\right)\)
\(=2\left(\frac{1}{18}+\frac{1}{12}+\frac{1}{9}\right)\)
\(\mathrm{LCM}\) of 18,12 and \(9=36\)
\(
=\frac{2(2+3+4)}{36}=\frac{1}{2}
\)
\(\therefore\) Remaining work \(=\frac{1}{2}\)
\(\therefore\) Required number of men \(=72 \times \frac{1}{2}=36\)
16 men can finish a piece of work in 49 days. 14 men started working and in 8 days they could finish certain amount of work. If it is required to finish the remaining work in 24 days. How many more men should be added to the existing workforce? [IBPS—RRB Office Assistant (Online) Exam, 2015]
Given
\(
\begin{aligned}
& M_1=16 ; \quad D_1=49 ; W_1=1 \\
& M_2=? ; \quad D_2=24 ; W_2=?
\end{aligned}
\)
According to the question
\(
\begin{aligned}
& \frac{M_1 D_1}{W_1}=\frac{M_2 D_2}{W_2} \\
& \Rightarrow \frac{16 \times 49}{1}=\frac{14 \times 8}{W_2} \\
& \Rightarrow W_2=\frac{14 \times 8}{16 \times 49}=\frac{1}{7}
\end{aligned}
\)
Remaining work \(=1-\frac{1}{7}=\frac{6}{7}\)
\(
\begin{aligned}
& \text { Again, } \frac{M_1 D_1}{W_1}=\frac{M_2 D_2}{W_2} \\
& \Rightarrow \frac{16 \times 49}{1}=\frac{M_2 \times 24}{\frac{6}{7}} \\
& \Rightarrow 16 \times 49=\frac{M_2 \times 24 \times 7}{6} \\
& \Rightarrow 16 \times 49=M_2 \times 4 \times 7 \\
& \Rightarrow M_2=\frac{16 \times 49}{4 \times 7}=28
\end{aligned}
\)
Number of additional men \(=28-14=14\)
A and B can do a work in 15 days, B and C in 30 days and A and C in 18 days. They work together for 9 days and then A left. In how many more days, can B and C finish the remaining work?
From the given information, we can have
\(
\begin{gathered}
(A+B) \text { ‘s } 1 \text { day work }=1 / 15 \\
(B+C) \text { ‘s } 1 \text { day work }=1 / 30 \\
(A+C) \text { ‘s } 1 \text { day work }=1 / 18 \\
(A+B+B+C+A+C) \text { ‘s } 1 \text { day work }=1 / 15+1 / 30+1 / 18 \\
(2 A+2 B+2 C) \text { ‘s } 1 \text { day work }=1 / 15+1 / 30+1 / 18 \\
2(A+B+C) ‘ s 1 \text { day work }=1 / 15+1 / 30+1 / 18
\end{gathered}
\)
L.C.M of \((15,30,18)=90\).
\(
\begin{gathered}
2(A+B+C) \text { ‘s } 1 \text { day work }=6 / 90+3 / 90+5 / 90 \\
2(A+B+C) \text { ‘s } 1 \text { day work }=14 / 90 \\
2(A+B+C) \text { ‘s } 1 \text { day work }=7 / 45 \\
(A+B+C) \text { ‘s } 1 \text { day work }=7 /(2 \cdot 45) \\
(A+B+C) \text { ‘s } 1 \text { day work }=7 / 90
\end{gathered}
\)
Then, the amount of work completed by \(A, B\) and \(C\) together in 9 days is
\(
\begin{aligned}
& =9 \cdot 7 / 90 \\
& =7 / 10
\end{aligned}
\)
Amount of work left for \(B\) and \(C\) to complete is
\(
=3 / 10
\)
Number of days that B will take to finish the work is
\(
\begin{aligned}
& =\text { amount of work/part of the work done in } 1 \text { day } \\
& =(3 / 10) /(1 / 30) \\
& =(3 / 10) \cdot(30 / 1) \\
& =9
\end{aligned}
\)
So, no. of days taken by \(B\) and \(C\) to finish the remaining work is 9 days.
A contractor decided to complete the work in 90 days and employed 50 men at the beginning and 20 men additionally after 20 days and got the work completed as per schedule. If he had not employed the additional men, how many extra days would he have needed to complete the work?
The work has to completed in 90 days (as per schedule).
Total no. of men appointed initially \(=50\).
Given : 50 men have already worked for 20 days and completed a part of the work.
If the remaining work is done by 70 men \((50+20=70)\), the work can be completed in 70 days and the total work can be completed in 90 days as per the schedule.
Let ‘ \(x\) ‘ be the no. of days required when the remaining work is done by 50 men.
For the remaining work,
\(
\begin{aligned}
& 70 \text { men } \cdots>70 \text { days } \\
& 50 \text { men } \cdots>\text { x days }
\end{aligned}
\)
The above one is an inverse variation.
Because, when no. of men is decreased, no. of days will be increased.
By inverse variation, we have
\(
\begin{aligned}
70 \cdot 70 & =50 \cdot x \\
4900 & =50 x \\
98 & =x
\end{aligned}
\)
So, if the remaining work is done by 50 men, it can be completed in 98 days.
So, extra days needed \(=98-70=28\) days.
Three taps A, B and C can fill a tank in 10, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, find the time taken to fill the tank.
From the given information, we can have
\(
\begin{aligned}
& \text { A’s } 1 \text { hour work }=1 / 10 \\
& \text { B’s } 1 \text { hour work }=1 / 15 \\
& C^{\prime} \text { s } 1 \text { hour work }=1 / 20
\end{aligned}
\)
In the first hour, we have
\(
\begin{gathered}
(A+B) \text { ‘s work }=1 / 10+1 / 15 =1 / 6 \\
\end{gathered}
\)
In the second hour, we have
\(
\begin{gathered}
(A+C) \text { ‘s work }=1 / 10+1 / 20 =3 / 20\\
\end{gathered}
\)
Amount of work done in each two hours is
\(
\begin{aligned}
& =1 / 6 +3 / 20 =19 / 60
\end{aligned}
\)
Amount of work done:
In the first 2 hours : \(19 / 60\)
In the first 4 hours : \(19 / 60+19 / 60=38 / 60\)
In the first 6 hours : \(19 / 60+19 / 60+19 / 60=57 / 60\)
After 6 hours, the remaining work will be
\(
\begin{aligned}
& =3 / 60 =1 / 20
\end{aligned}
\)
\(1 / 20\) is the small amount of work left and \(A\) alone can complete this.
Time taken by \(A\) to complete this \(1 / 20\) part of the work is
\(=\) amount of work/part of work done in 1 hour
\(
\begin{aligned}
& =(1 / 20) /(1 / 10) \\
& =(1 / 20) \cdot(10 / 1) \\
& =1 / 2 \text { hours }
\end{aligned}
\)
So, A will will take half an hour (or 30 minutes) to complete the remaining work \(1 / 20\).
So, the total time taken to complete the work is
\(=6\) hours +30 minutes
\(=6 \frac{1}{2}\) hours
 A builder appoints three construction workers Akash, Sunil and Rakesh on one of his sites. They take 20, 30 and 60 days respectively to do a piece of work. How many days will it take Akash to complete the entire work if he is assisted by Sunil and Rakesh every third day?
Total work done by Akash, Sunil and Rakesh in 1 day \(=\{(1 / 20)+(1 / 30)+(1 / 60)\}=1 / 10\)
Work done along by Akash in 2 days \(=(1 / 20) \times 2=1 / 10\)
Work Done in 3 days ( 1 day of all three together +2 days of Akash’s work) \(=(1 / 10)+(1 / 10)=1 / 5\)
So, work done in 3 days \(=1 / 5\)
Time taken to complete the work \(=5 \times 3=15\) days
To complete a piece of work, Samir takes 6 days and Tanvir takes 8 days alone respectively. Samir and Tanvir took Rs.2400 to do this work. When Amir joined them, the work was done in 3 days. What amount was paid to Amir?
Total work done by Samir and Tanvir \(=\{(1 / 6)+(1 / 8)\}=7 / 24\)
Work done by Amir in 1 day \(=(1 / 3)-(7 / 24)=1 / 24\)
Amount distributed between each of them \(=(1 / 6):(1 / 8):(1 / 24)=4: 3: 1\)
Amount paid to Amir \(=(1 / 24) \times 3 \times 2400=\) Rs. 300
Dev completed the school project in 20 days. How many days will Arun take to complete the same work if he is \(25 \%\) more efficient than Dev?
Let the days taken by Arun to complete the work be \(x\)
The ratio of time taken by Arun and \(D e v=125: 100=5: 4\)
\(
\begin{aligned}
& 5: 4: 20: x \\
& \Rightarrow x=\{(4 \times 20) / 5\} \\
& \Rightarrow x=16
\end{aligned}
\)
Time taken by A to finish a piece of work is twice the time taken B and thrice the time taken by C. If all three of them work together, it takes them 2 days to complete the entire work. How much work was done by B alone?
Time taken by \(A=x\) days
Time taken by \(B=x / 2\) days
Time Taken by \(\mathrm{C}=\mathrm{x} / 3\) days
\(
\begin{aligned}
& \Rightarrow\{(1 / x)+(2 / x)+(3 / x)=1 / 2 \\
& \Rightarrow 6 / x=1 / 2 \\
& \Rightarrow x=12
\end{aligned}
\)
Time taken by \(B=x / 2=12 / 2=6\) days
Sonal and Preeti started working on a project and they can complete the project in 30 days. Sonal worked for 16 days and Preeti completed the remaining work in 44 days. How many days would Preeti have taken to complete the entire project all by herself?
Let the work done by Sonal in 1 day be \(x\)
Let the work done by Preeti in 1 day be \(y\)
Then, \(x+y=1 / 30 \dots(i)\)
\(
\Rightarrow 16 x+44 y=1 \dots(ii)
\)
Solving equation (i) and (ii),
\(
\begin{aligned}
& x=1 / 60 \\
& y=1 / 60
\end{aligned}
\)
Thus, Preeti can complete the entire work in 60 days
In a factory, 20 people can make 20 toys in 15 days working 10 hours per day. Then, in how many days can 25 persons make 30 toys working 20 hr per day?
Here, \(M 1=20, M 2=25, D 1=15, D 2=?, T 1=10, T 2=20, W 1=20\) and \(W 2=30\)
We know,
\(
\begin{aligned}
& M 1 \times D 1 \times T 1 \times W 2=M 2 \times D 2 \times T 2 \times W 1 \\
& \Rightarrow 20 \times 15 \times 10 \times 30=25 \times D 2 \times 20 \times 20 \\
& \Rightarrow D 2=9 .
\end{aligned}
\)
Thus, the required day \(=9\) days
\(P, Q\) and \(R\) together can complete a work in 16 days and \(R\) alone complete the work in 20 days. If \(P, Q\) and \(R\) started the work together and after 10 days \(P\) and \(Q\) left the work, in how many days \(\mathrm{R}\) alone complete the remaining work?
\(
\begin{aligned}
& P+Q+R=16 \text { days } \\
& R=20 \text { days } \\
& \text { Total work (LCM of } 16 \text { and } 20)=80 \\
& (P+Q+R) \text { ‘s work } \\
& =80 / 16 \\
& =5 \text { unit } \\
& \text { Work done by } R=80 / 20=4 \text { unit } \\
& (P+Q+R)^{\prime} \text { s } 10 \text { days work } \\
& =5 \times 10 \\
& =50 \text { unit } \\
& \text { Remaining work } \\
& =(80-50) \\
& =30 \text { unit } \\
& \text { Remaining work done by } R \\
& =30 / 4 \\
& =7 \frac{1}{2} \text { days }
\end{aligned}
\)
Two brothers(younger and older) can do a piece of work in 70 and 60 days respectively. They began to work together, but the younger brother leaves after some days and the older brother finished the remaining work in 47 days. After how many days did the younger brother leave?
The Older brother would have done \(47 / 60\) work in 47 days.
The remaining work \(=(1-47 / 60)=13 / 60\) must have done by them together.
Two brothers 1 day work \(=(1 / 70+1 / 60)=13 / 420\)
So, they would have done \(13 / 60\) work in \(420 / 13 \times 13 / 60=7\) days.
Therefore, The younger brother left the work after 7 days.
Brij alone takes 3 days more than Brij and Mohan together to complete work while Mohan takes 12 days more than Brij and Mohan together, then in how many days Brij and Mohan together can complete the work?
Here, \(d_1=3\) and \(d_2=12\)
So, the Time taken by Brij and Mohan to complete the work together will be \(=\sqrt{3 \times 12}=\sqrt{36}=6\) days
8 men and 7 women can complete a piece of work in 15 days while 6 men and 3 women can complete a piece of work in 30 days. In how many days 1 woman can complete the total work?
Let us assume efficiencies or one day work of 1 man and 1 woman are \(\mathrm{M}\) and \(\mathrm{W}\) respectively
So, the efficiency of 8 men \(=8 \mathrm{M}\) and efficiency of 7 women \(=7 \mathrm{~W}\)
Similarly, efficiency of \(6 \mathrm{men}=6 \mathrm{M}\) and efficiency of 3 women \(=3 \mathrm{~W}\)
We know, Work done \(=\) time taken \(\times\) Efficiency
According to question, 8 men and 7 women can complete a piece of work in 15 days
So, Total work done \(=(\) Efficiency of 8 men and 7 women \() \times\) Time taken
\(\Rightarrow\) Total work done \(=(8 \mathrm{M}+7 \mathrm{~W}) \times 15=120 \mathrm{M}+105 \mathrm{~W} \dots(1)\)
Similarly, 6 men and 3 women can complete a piece of work in 30 days
So, Total work done \(=(\) Efficiency of 6 men and 3 women \() \times\) Time taken
\(\Rightarrow\) Total work done \(=(6 M+3 W) \times 30=180 M+90 W \dots(2)\)
As, total work done is same so
\(
\begin{aligned}
& 120 M+105 W=180 M+90 W \\
& 60 M=15 W \\
& 4 M=1 W
\end{aligned}
\)
From the above equation we can say that efficiency of 1 woman is equal to the efficiency of 4 men
Now, from equation (1)
Total work done \(=120 \mathrm{M}+105 \mathrm{~W}\)
\(\Rightarrow\) Total work done \(=\frac{120}{4} W+105 W=135 W\)
Therefore, time taken by 1 woman to complete the work will be \(\Rightarrow \frac{135 W}{1 W}=135\) days
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